\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
deepdream of
          a sidewalk
Show Question
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\groupAdd}[1] { +_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \require{physics} \require{ams} \require{mathtools} \)
Math and science::INF ML AI

Inverse of symmetric matrix is symmetric

The inverse of a symmetric matrix is symmetric.

The proof is on the reverse side. Can you remember it?

There is an intuitive perspective that relies on diagonalizing the matrix.


Proof

Let \( A \) be a nonsingular symmetric matrix. We want to show that \( A^{-1} \) is symmetric.

To start with, we know that:

\[ \begin{align*} I &= A A^{-1} \\ &= (A A^{-1})^T && \text{as $I = I^{T}$} \\ &= (A^{-1})^T A^T && \text{by the rule $(AB)^{T} = B^TA^T$} \\ &= (A^{-1})^T A && \text{as $A = A^T$} \\ \end{align*} \]

So, \( (A^{-1})^T \) is also an inverse of \( A \), but there is only one, so \( A^{-1} = (A^{-1})^T \).

Symmetric matrix has orthogonal eigenvectors

The Spectral theorem tells us that a symmetric matrix has orthogonal eigenvectors and can be decomposed into a rotation, a scaling and a rotation back. To undo this opperation, we would again carry out the same rotation, then reverse the scaling and rotate back. This inversing operation has exactly the same form and is also symmetric.

Symbolically, a symmetric matrix \( A \) can be decomposed as:

\[ A = Q \Delta Q^T  \]

where \( Q \) is an orthogonal matrix of eigenvectors of \( A \) and \( \Delta \) is a diagonal matrix containing the eigenvalues.

Taking the transpose of this representation, we get:

\[ (Q \Delta Q^T)^T = Q \Delta^T Q^T = Q \Delta Q^T \\ \]

showing that \( Q \Delta Q^T \) (and thus \( A \)) is symmetric.